Level 3C: Differential Forms – The Exterior Derivative

The Defect. In multivariable calculus, the objects that integrate are somehow “infinitesimal products of differentials”: $dx$, $dx\,dy$, and the cross product of normals. The rules for manipulating them are memorized as formal moves, with no account of what the symbols denote. Differential forms give the symbols referents: a $k$-form is a machine that eats $k$ vectors and returns a signed $k$-dimensional volume, and every formal rule becomes a theorem. This page builds the algebra; the next integrates it and proves Stokes.

The One-Line Reveal. A differential form is a determinant evaluator: $dx^I$ applied to $k$ vectors returns the determinant of their $I$-coordinates, and everything (wedge products, the exterior derivative, $d^2 = 0$, coordinate independence) follows from properties of determinants.

Scope. From Level 1B: determinants as alternating forms (Section 4), exterior powers and their dimension (Theorem 5.4), the dual basis (Section 2). From Level 3A: smooth manifolds, charts, smooth maps, the algebra of derivatives. From Level 3B: the chain rule in chart form. One input is declared: Clairaut’s theorem (if the second partial derivatives of a smooth function are continuous, they are equal; proved in the Real Analysis track). It is used in exactly two places, both flagged. Everything else is proved.

Definition. An alternating $k$-form at a point $a \in \mathbb{R}^n$ is an alternating $k$-linear map $\omega: (\mathbb{R}^n)^k \to \mathbb{R}$ (1B, Section 4): linear in each argument, zero when two arguments coincide. For $k = 0$ this is a scalar. A $k$-form field on an open set $U \subseteq \mathbb{R}^n$ is an assignment $a \mapsto \omega(a)$ of an alternating $k$-form at each $a$, smooth in $a$ in the sense made precise below.

Definition (the basis forms). For an index sequence $I = (i_1, \dots, i_k)$ with entries in $\{1, \dots, n\}$, define the alternating $k$-form $dx^I$ by the determinant formula (1B, Section 4):

the determinant of the matrix whose $(p, q)$ entry is the $i_p$-th coordinate of $v_q$. The entries $x^i$ are the coordinate functionals, the dual basis of the standard basis (1B, Theorem 2.1). Since matrices with two equal rows have determinant zero and transposing rows negates the determinant, each $dx^I$ is alternating. Write $dx^i := dx^{(i)}$, so $dx^i(v) = v_i$.

Lemma 1.1 (insertion rules). For an index sequence $I$ of length $k$ and an index $j$: if $j$ occurs in $I$, then $dx^j \wedge dx^I = 0$, where the wedge is defined below; if $j$ does not occur in $I$, then

where $I \oplus j$ is the increasing rearrangement of the sequence $(j, i_1, \dots, i_k)$ and $\operatorname{sgn}(\sigma_I^j)$ is the sign of the permutation that performs that rearrangement (Proposition 4.3 of 1B makes the sign well defined).

Proof. The wedge is defined first on the basis symbols: for sequences $I$ of length $k$ and $J$ of length $l$,

the same determinant formula applied to the concatenated index sequence $I \concat J$. (Repeated indices make some two rows of the matrix equal, so the determinant vanishes: the first half of Lemma 1.1 with $j \in I$ is exactly this.) For the sign computation with $j \notin I$: both $\operatorname{sgn}(\sigma_I^j)\, dx^{I \oplus j}$ and $dx^j \wedge dx^I$ are alternating $(k{+}1)$-forms, and Proposition 4.6 of 1B (determination of alternating forms by their values on increasing tuples) says it suffices to compare their values on the standard basis vectors. For an increasing $(k{+}1)$-tuple $K$, the matrix $\big(x^{(j \concat I)_p}(e_{K_q})\big)$ is a permutation matrix (its rows select distinct standard coordinates), and its determinant is the sign of the row-permutation carrying $(j \concat I)$ to the increasing order, which is exactly $\operatorname{sgn}(\sigma_I^j)$ when $K = I \oplus j$, and the matrix has a zero row otherwise. Both sides of the claimed identity therefore agree on every increasing tuple of standard basis vectors, and values on increasing tuples determine an alternating form (Proposition 1.2 below), so the forms are equal. $\blacksquare$

Proposition 1.2 (coordinate basis). Every alternating $k$-form $\omega$ at a point of $\mathbb{R}^n$ has a unique expression

where $e_I = (e_{i_1}, \dots, e_{i_k})$ is the tuple of standard basis vectors indexed by the increasing sequence $I$. Consequently the increasing $dx^I$ form a basis of the space of alternating $k$-forms, and that space has dimension $\binom{n}{k}$, matching 1B’s exterior power dimension (Theorem 5.4).

Proof. Evaluation. For increasing $I$ and $J$: $dx^J(e_I) = \det\big(\delta_{(J)_p, (I)_q}\big)$, which is $1$ if $J = I$ (the identity matrix) and $0$ otherwise (the index sets differing, some row is zero).

Spanning. Expand each argument $v_q = \sum_i v_{i,q}\, e_i$ inside $\omega(v_1, \dots, v_k)$ by multilinearity. Terms with a repeated basis vector vanish by alternation. Terms with distinct indices sort into increasing order by transpositions of arguments, each transposition negating the value: $0 = \omega(\dots, v{+}w, \dots, v{+}w, \dots)$ expands to $\omega(v, w) + \omega(w, v)$ (the argument of 1B’s Lemma 4.4), so swapping two arguments negates. The collected sign is $(-1)^m$ for the sorting permutation, well defined by 1B’s Proposition 4.3(i). Collecting terms gives

Independence. If $\sum_I a_I\, dx^I = 0$, evaluate at $e_J$: $a_J = 0$. $\blacksquare$

Definition (wedge of fields). For each $k$ and $l$, the collections of alternating $k$- and $l$-forms at a point are vector spaces (pointwise operations), and by Proposition 1.2 every alternating $k$-form at a point has a unique expression $\sum_I a_I\, dx^I$ over increasing $I$. For fields $\omega = \sum_I a_I\,dx^I$ and $\eta = \sum_J b_J\,dx^J$ (finitely supported coefficient functions), we define

Uniqueness of the basis expression makes the definition legitimate (the same field cannot be expanded two ways and receive two wedges).

Proposition 1.3 (the wedge is associative and graded commutative). The wedge of fields is bilinear; $(\omega \wedge \eta) \wedge \theta = \omega \wedge (\eta \wedge \theta)$; and $\eta \wedge \omega = (-1)^{kl}\, \omega \wedge \eta$ for a $k$-form $\omega$ and an $l$-form $\eta$. In particular transposing $dx^I$ past $l$ factors of $dx^J$ contributes the sign $(-1)^{kl}$.

Proof. Bilinearity is bilinearity of the coefficients in the definition. Associativity: at the level of basis symbols, both $(dx^I \wedge dx^J) \wedge dx^K$ and $dx^I \wedge (dx^J \wedge dx^K)$ are, by the definition, the determinant formula applied to the single concatenated sequence $I \concat J \concat K$; so they are equal as alternating maps, and the general claim follows by expanding the fields in their bases (Proposition 1.2) and using bilinearity. Graded commutativity: it suffices to check basis symbols $dx^I \wedge dx^J = (-1)^{kl} dx^J \wedge dx^I$. Both sides are alternating $(k{+}l)$-forms that vanish unless $I \cap J = \emptyset$ (Lemma 1.1, applied repeatedly). When $I \cap J = \emptyset$, evaluation on the increasing tuple $(e_1, \dots, e_{k+l})$ gives, on both sides, the determinant of a permutation matrix: the left side carries the permutation sorting $I \concat J$ into increasing order, the right side the permutation sorting $J \concat I$. These two permutations differ by a reversal of $k$ elements past $l$ elements, and the sign of that reversal is $(-1)^{kl}$ (it is the product of $kl$ adjacent transpositions, each of sign $-1$). By Proposition 1.2 (values on increasing tuples determine), the alternating forms agree. $\blacksquare$

Definition (smoothness of fields). A $k$-form field $\omega = \sum_I a_I\, dx^I$ on $U$ is smooth if the coefficient functions $a_I$ are smooth, and this does not depend on the choice of basis expansion (the coefficients are unique at every point, so there is nothing to check). Write $\Omega^k(U)$ for the vector space of smooth $k$-form fields on $U$, and $\Omega^0(U) = C^\infty(U)$.

Quick check.

1. Evaluate dx^I on k vectors.
2. When is dx^j wedge dx^I zero?
3. What is dx^j wedge dx^I when j is not in I?
4. Why do the increasing dx^I form a basis?
5. What is the wedge of two fields?

Answers: 1. The determinant of the matrix of their I-indexed coordinates. 2. When j occurs in I (a repeated row). 3. The sorting sign times dx of the merged increasing sequence (Lemma 1.1). 4. Evaluation on increasing tuples extracts unique coefficients (Proposition 1.2). 5. Pairwise coefficient products times the basis wedges.

Definition. For a smooth function $a \in C^\infty(U)$, define

For $\omega = \sum_I a_I\, dx^I \in \Omega^k(U)$, define

which is a $(k{+}1)$-form field by the insertion rules (Lemma 1.1). The operator $d: \Omega^k(U) \to \Omega^{k+1}(U)$ is the exterior derivative. On functions it reproduces Section 1’s definition.

Lemma 2.1 (the product rule). For smooth $a, b$: $d(ab) = a\, db + b\, da$.

Proof. $d(ab) = \sum_i \partial_i(ab), dx^i = \sum_i (\partial_i a), b

• a, (\partial_i b), dx^i = b, da + a, db$(ordinary product rule), and scalars pull through the wedge.$\blacksquare$

Theorem 2.2 ($d^2 = 0$). For every smooth $k$-form field $\omega$, we have $d(d\omega) = 0$.

Proof. By linearity of $d$ in the coefficients, it suffices to check $d(da) = 0$ for a function $a$, since

the exterior derivative acting only on the coefficient functions. Now $da = \sum_i \partial_i a\, dx^i$, so

The diagonal terms vanish ($dx^i \wedge dx^i = 0$, Lemma 1.1). The off-diagonal terms pair up: for $i \neq j$,

using $dx^i \wedge dx^j = -\, dx^j \wedge dx^i$ (Proposition 1.2 with $k = l = 1$) and Clairaut’s theorem (the declared input) to equate the mixed partials. $\blacksquare$

Theorem 2.3 (the graded Leibniz rule). For a $k$-form $\omega$ and an $l$-form $\eta$:

Proof. By bilinearity, expand in the bases: $\omega = \sum_I a_I dx^I$, $\eta = \sum_J b_J dx^J$. Then $\omega \wedge \eta = \sum_{I,J} a_I b_J\, dx^I \wedge dx^J$, and

by Lemma 2.1. The first summand reorganizes by associativity (Proposition 1.2):

The second summand needs $db_J$ carried past the $k$ factors of $dx^I$. Transposing a 1-form past a single alternating factor negates (Proposition 1.2, graded commutativity with $k = l = 1$), so transposing past $k$ of them contributes $(-1)^k$:

and summing gives $(-1)^k\, \omega \wedge d\eta$. $\blacksquare$

Quick check.

1. Compute d of x squared times y.
2. Why is d squared zero on functions?
3. State the graded Leibniz rule.
4. Verify d of df vanishes for f = x squared y in coordinates.
5. What is d of (a dx^I)?

Answers: 1. 2xy dx plus x squared dy. 2. Mixed partials agree, and the antisymmetric pairing cancels the terms (Theorem 2.2, Clairaut). 3. d of a wedge = d of the first wedge the second, plus minus one to the degree of the first, times the first wedge d of the second (Theorem 2.3). 4. The dy-wedge-dx term and the dx-wedge-dy term cancel by antisymmetry. 5. Sum over j of the j-th partial times dx^j wedge dx^I.

Definition (pullback). For a smooth map $f: U \to V$ between open sets of Euclidean spaces and $\omega = \sum_J b_J\, dy^J \in \Omega^k(V)$, the pullback is

where $f^{j} = y^j \circ f$ are the component functions and $df^j = \sum_i \partial_i f^j\, dx^i$; the wedge is associativity-consistent (Proposition 1.2). For functions, $f^*b = b \circ f$. Pullback is linear in $\omega$ and contravariantly compositional: $(g \circ f)^* = f^* \circ g^*$ (both sides substitute $g(f(x))$ for $y$; a one-line pointwise check).

Theorem 3.1 ($d$ commutes with pullback). For every $k$-form $\omega$ on $V$: $d(f^*\omega) = f^*(d\omega)$.

Proof. By linearity, expand $\omega = \sum_J b_J\, dy^{j_1} \wedge \cdots \wedge dy^{j_k}$ over increasing $J$. First,

Second, by the graded Leibniz rule (Theorem 2.3) applied successively to the product $(b_J \circ f)\, df^{j_1} \wedge \cdots \wedge df^{j_k}$, whose first factor is a function:

Every term of the second sum vanishes: $d(df^{j_s}) = d\big(\sum_i \partial_i f^{j_s}\, dx^i\big) = \sum_{i, r} \partial_r \partial_i f^{j_s}\, dx^r \wedge dx^i = 0$ by $d^2 = 0$ on functions (Theorem 2.2, Clairaut again). For the first sum, the chain rule gives $\partial_i (b_J \circ f) = \sum_l (\partial_l b_J \circ f)\, \partial_i f^l$, so

and substituting shows $d(f^*\omega)$ equals the display for $f^*(d\omega)$. $\blacksquare$

Theorem 3.2 ($d$ is coordinate independent on manifolds). Let $M$ be a smooth manifold. A $k$-form field on $M$ is a collection of $k$-form fields $\omega_\varphi$ on $\varphi(U)$, one for each admissible chart $(U, \varphi)$, such that on overlaps the expressions correspond through the transition maps: $\omega_\psi = (\psi \circ \varphi^{-1})^*\, \omega_\varphi$ wherever both are defined. Then the charts’ exterior derivatives assemble: there is a unique operator $d_M: \Omega^k(M) \to \Omega^{k+1}(M)$ with $(d_M\omega)_\psi = d(\omega_\psi)$ in every chart; and it agrees with the Euclidean $d$ on chart representations.

Proof. Consistency is exactly Theorem 3.1. Let $T = \psi \circ \varphi^{-1}$ be a transition. The compatibility condition gives $\omega_\psi = T^* \omega_\varphi$ on $\psi(U \cap V)$. Applying $d$:

by Theorem 3.1. So the $d$ of the $\psi$-representation equals the pullback of the $d$ of the $\varphi$-representation, which is precisely the compatibility condition for the family $\{d(\omega_\varphi)\}$ to be a $(k{+}1)$-form field on $M$. Existence: on a chart domain, set $(d_M\omega)_\varphi := d(\omega_\varphi)$; the computation just shown says the pieces agree on overlaps, so they assemble into a globally defined field. Uniqueness: any operator agreeing with the Euclidean $d$ on every chart representation is pinned down on every chart domain, and chart domains cover $M$. Independence of the atlas: admissible charts all belong to one maximal atlas, and the pieces were checked across arbitrary transitions between admissible charts. $\blacksquare$

Remark (what $d$ is good for). On functions, $df_p$ from Level 3B evaluates on a tangent vector $X$ as $X([f])$; the field $df$ of Theorem 3.2 is that object, varying smoothly over $M$. On $n$-forms in $n$-dimensional space, $d$ produces exactly the density that the next page integrates: applied to $\omega = \sum_i (-1)^{i-1} F_i\, dx^1 \wedge \cdots \widehat{dx^i} \cdots \wedge dx^n$, it yields $(\operatorname{div} F)\, dx^1 \wedge \cdots \wedge dx^n$, a computation included as a self-test.

Quick check.

1. Define the pullback of b dy^J.
2. Which proved rule kills the d-of-df terms in the pullback proof?
3. Which rule rewrites d of the pulled-back coefficient?
4. State the coordinate-independence theorem for d on manifolds.
5. What is the pullback of (g times omega)?

Answers: 1. The pulled-back coefficient times the wedge of the pulled-back differentials. 2. d squared equals zero on functions (Theorem 2.2). 3. The chain rule (Theorem 3.1’s first sum). 4. The chart-wise d satisfies the compatibility condition because d commutes with transitions (Theorem 3.2). 5. (g composed with f) times the pullback of omega.

1. Prove that $dx^I = 0$ as an alternating map whenever $I$ has a repeated index. (§1 – two rows of the determinant matrix coincide.)
2. Prove that the increasing families $\{dx^I\}$ are linearly independent as alternating maps, citing the exact theorem of Level 1B you use. (§1 – 1B, Theorem 5.4: the exterior power has basis the increasing wedges; the $dx^I$ are that basis read through the duality of alternating maps and alternating multilinear forms.)
3. Compute $d(\omega \wedge \omega)$ for a 1-form $\omega$ two ways: by the graded Leibniz rule and by expanding $\omega$ in coordinates; confirm both give $0$. (§2 – $d\omega \wedge \omega - \omega \wedge d\omega$ and the fact that $\omega \wedge \omega = 0$ for odd-degree forms.)
4. Prove the claim of the Remark: for $F = (F_1, \dots, F_n)$ smooth and $\omega = \sum_i (-1)^{i-1} F_i\, dx^1 \wedge \cdots \widehat{dx^i} \cdots \wedge dx^n$, one has $d\omega = (\operatorname{div} F)\, dx^1 \wedge \cdots \wedge dx^n$. (§2, §1 – apply $d$ term by term; each $dF_i$ contributes $\partial_i F_i\, dx^i$ in the slot left open, and Lemma 1.1’s insertion rule sorts the sign.)
5. Show that $d$ of a 0-form in one chart, read through a transition map, matches $d$ in the other chart, for $M = \mathbb{R}$ with charts $x$ and $y = x^3$. (§3 – $d_x$ of $a$ versus $d_y$ of $a \circ y^{-1}$; the chain rule closes the gap.)
6. Prove that if $\omega$ is closed ($d\omega = 0$) then so is every pullback $f^*\omega$. (§3, §2 – $d(f^*\omega) = f^*(d\omega) = f^*0 = 0$.)

Problems. The quick checks above tested recall. These chain several results and require full written arguments; hints are included where the jump is large.

7. Compute $d(x^2 y\, dx)$ and verify $d^2$ vanishes on it by hand. ($d(x^2y\,dx) = 2xy\, dy \wedge dx$; its $d$ pairs $2x\,dx \wedge dy \wedge dx = 0$ with the $dy$-term vanishing by repetition.)
8. Prove $f^*(dx^1 \wedge dx^2) = \det(dF)\, dx^1 \wedge dx^2$ for a smooth $F: \mathbb{R}^2 \to \mathbb{R}^2$. (Expand $df^1 \wedge df^2$ by the insertion rules; the cross terms die and the survivor is the two-by-two determinant.)
9. Prove that pullback is contravariant: $(g \circ f)^* = f^* \circ g^*$. (A pointwise substitution chase through the definition.)

The entire page follows from one construction: forms are determinant evaluators, and $d$ is the unique operator that differentiates their coefficients antisymmetrically. Determinants give alternation for free (associativity of the wedge becomes definitional), Clairaut’s theorem kills $d^2$, and the chain rule converts the two into the statement that $d$ means the same thing in every coordinate system.

• Level 3D: Integration and Stokes – orientation, integration of forms, and the generalised Stokes theorem, where the exterior derivative of this page starts paying rent.
• Level 3B: Tangent Spaces – the differential whose field-version is the $df$ of this page.
• An Introduction to Manifolds – Loring Tu’s Chapters 3 and 6, which develop the exterior derivative and its coordinate independence with the same determinant-first emphasis.